Subtract the following rational expressions. $\dfrac{4k+23}{k^2-15k}-\dfrac{k^2+6k}{k^2-15k}=$
Solution: We want to subtract two rational expressions whose denominators are equal. We can do this by subtracting the numerators and keeping the denominator the same. [Does this fit with how we subtract rational numbers?] $\begin{aligned} &\phantom{=}\dfrac{4k+23}{k^2-15k}-\dfrac{k^2+6k}{k^2-15k} \\\\ &=\dfrac{(4k+23)-(k^2+6k)}{k^2-15k} \\\\ &=\dfrac{4k+23-k^2-6k}{k^2-15k} \\\\ &=\dfrac{-k^2-2k+23}{k^2-15k} \end{aligned}$ In conclusion, $\dfrac{4k+23}{k^2-15k}-\dfrac{k^2+6k}{k^2-15k}=\dfrac{-k^2-2k+23}{k^2-15k}$